Prove : $\forall x>0$, $\arctan(x)>\frac{x}{1+x^2}$ I think the question is itself wrong as, $\frac{x}{1+x^2}$ is the derivative of $\arctan x$. If not, I don't know how to proceed.
 A: Hint
If
$$g(x)=\arctan(x)-\frac{x}{1+x^2},$$
then $g'(x)>0$ for all $x>0$. Therefore $g(x)>g(0)$ for all $x>0$.
A: Letting $x=\tan\theta$ for some $\theta\in\left(-\pi/2,\pi/2\right)$ we are left with
$$ 2\theta > \sin(2\theta) $$
which holds for any $\theta\in(0,\pi/2)$, by convexity.
A: Hint. Consider the derivative of $$\arctan x-\frac{x}{1+x^2},$$ namely $$\frac{2x}{(1+x^2)^2},$$ which is positive for all $x>0.$ Thus, the difference is increasing as $x\to+\infty.$ Since the difference is $0$ when $x=0,$ it follows that the difference is positive for all $x>0,$ and the result immediately follows.
A: Let $$f(x)=\tan^{-1} x-\frac{x}{1+x^2} \implies f'(x)=\frac{2x^2}{(1+x^2)^2}>0$$
so $f(x)$is an increasing function for $x \in (0,\infty)$. Then $$f(x) \ge f(0)=0 \implies \tan^{-1}x \ge \frac{x}{1+x^2}.$$
A: Consider the function $$f(t) =\arctan t-\frac{t}{1+x^2}$$ on interval $[0,x]$. Clearly $f'(t) >0 $ for all $t\in(0,x)$ and hence $f$ is strictly increasing on $[0,x]$ so that $f(x) >f(0)=0$ ie $$\arctan x-\frac{x} {1+x^2}>0$$

You will find that the form of $f(t) $ is such that taking derivatives is very easy and checking sign of derivative is also easy.
Another equivalent approach is via mean value theorem. We have $$\arctan x =\arctan 0+x\cdot\frac{1}{1+c^2}$$ for some $c\in(0,x)$. And since $c<x$ we have $$\frac{1}{1+c^2}>\frac{1}{1+x^2}$$ so that $$\arctan x>\frac{x} {1+x^2}$$ for all $x>0$.
